Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in PYTHON):

Solution:
import sys
from collections import deque
import copy
import math
def get_read_func(fileobject):
    if fileobject == None :
        return raw_input
    else:
        return fileobject.readline

def main():
    if len(sys.argv) > 1:
        f = open(sys.argv[1])
    else:
        f = None
    read_func = get_read_func(f);
    input_raw = read_func().strip().split()
    [N] = [int(input_raw[0])]
    P = []
    input_raw = read_func().strip().split()
    for i in range(N):
        P.append(float(input_raw[i]))

    dp = [[0 for i in range(N + 1)] for j in range(N + 1)]
    dp[0][0] = 1
    for i in range(0, N):
        for j in range(0, i + 1):
            dp[i + 1][j] += (1.0 - P[i]) * dp[i][j]
            dp[i + 1][j + 1] += P[i] * dp[i][j]

            #dp[i][j] += p[i] * dp[i - 1][j - 1] + (1 - p[i])

    count = 0L
    for j in range(N/2 + N%2, N + 1):
        count += dp[N][j]
    print count
if __name__ == '__main__':
    main()